Explanation

(b) As is known, frequency of vibration of a stretched string
$n \propto \sqrt{T} \propto \sqrt{mg} \propto \sqrt{g}$
As$n_{\omega} = \frac{80}{100}n_{a} = 0.8n_{a}$
$\therefore\frac{g^{'}}{g} = \left( \frac{n_{\omega}}{n_{a}} \right)^{2} = (0.8)^{2} = 0.64$
If $\rho_{\omega} =$ relative density of water(=1)
$\rho_{m} =$relative density of mass
$\rho_{r} =$ relative density of liquid, then
$\frac{g^{'}}{g} = \left( 1 - \frac{\rho_{\omega}}{\rho_{m}} \right) = 0.64$
$\frac{\rho_{\omega}}{\rho_{m}} = 1 - 0.64 = .36$ (i)
Similarly, in the liquid
$\frac{g^{'}}{g} = \left( \frac{n_{L}}{n_{a}} \right)^{2} = (0.6)^{2} = 0.36$
$\frac{g^{'}}{g} = \left( 1 - \frac{\rho_{L}}{\rho_{m}} \right) = 0.36$
$\frac{\rho_{L}}{\rho_{m}} = 1 - 0.36 = 0.64$ (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{\rho_{L}}{\rho_{m}} = \frac{0.64}{0.34} = 1.77$
Hence specific gravity of liquid=1.77